Trippy Triangles

Geometry Level 3

In how many ways can we form a nondegenerate triangle by choosing three distinct numbers from the set {1, 2, 3, 4, 5} as the sides?

Image Credit: Flickr Arnaud Huc .


The answer is 3.

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1 solution

Tijmen Veltman
Aug 29, 2015

We need our three numbers a , b , c a,b,c to satisfy the triangular inequalities:

a + b > c a+b>c

b + c > a b+c>a

c + a > b c+a>b

We can see that none of the numbers can be equal to 1 1 , since this would infer (setting a = 1 a=1 ) that b + 1 > c b+1>c and c + 1 > b c+1>b which is impossible for distinct integers b , c b,c . The remaining combinations to check are:

( 2 , 3 , 4 ) (2,3,4) : valid;

( 2 , 3 , 5 ) (2,3,5) : invalid (this gives you a degenerate triangle, since 2 + 3 = 5 2+3=5 );

( 2 , 4 , 5 ) (2,4,5) : valid;

( 3 , 4 , 5 ) (3,4,5) : valid.

Hence we are left with 3 \boxed{3} possible triangles.

What about the equilateral triangles, 1,1,1; 2.2.2; 3.3.3; 4.4.4; 5.5.5 ? What about all the isosceles triangles, 2,2,1; 3,3,1; 4,4,1....etc. I'm sure there are more than 3 triangles!

Gwen Roberts - 5 years, 9 months ago

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Because the problem specifies distinct numbers for the sides, none of those qualify.

Jason Short - 5 years, 9 months ago

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Ahhhh, I missed that!

Gwen Roberts - 5 years, 9 months ago

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